Taylor Series Theorem

Complex Analysis · Axiom Academy

LESSON Taylor's Theorem for Analytic Functions In the complex plane, being differentiable once forces a function to equal its own Taylor series. Let f be analytic on the open disk . Taylor's theorem guarantees a single power series in (z-z_0) that converges to f at every point of that disk, and the coefficients are forced to be a_n = f^ (n) (z_0)/n! — exactly the familiar Taylor formula, now with no exceptions. the analytic function, on the disk the coefficients are pinned by the derivatives at z_0 How big is the disk? Grow it outward from z_0 until its edge first reaches a point where f stops being analytic. The radius of convergence R is exactly that distance — the gap from the center to the nearest singularity. Why does the series equal the function? The proof feeds Cauchy's integral formula a geometric-series expansion of 1/(w-z) and reads off the coefficients. The payoff is concrete: inside the disk the partial sums march steadily onto f , and the error vanishes as . Valid exactly when , i.e. for z inside the contour. Substituting into and swapping sum with integral hands you the Taylor coefficients directly. Take the model case expanded at z_0=0 . Its only singularity is at z=1 , so R=1 . Watch the partial sums of converge to f across the disk and run off to the asymptote at the edge.

This is the written version of the interactive lesson above. See the full Complex Analysis course.